Problem 86 - Intermediate Algebra

Question

Perform the operations and simplify. $$ \frac{x^{2}+3 x+x y+3 y}{x^{2}-9} \cdot \frac{3-x}{x^{3}+3 x^{2}} $$

Step-by-Step Solution

Answer

The expression simplifies to \(\frac{-(x + y)}{x^2}\).

Step 1: Factor the Numerator and Denominator

First, factor both the numerator and denominator of the two fractions. In the first fraction, the numerator is \(x^{2} + 3x + xy + 3y\). Factor by grouping: \(x(x + 3) + y(x + 3) = (x + y)(x + 3)\). The denominator \(x^2 - 9\) can be factored as a difference of squares: \((x - 3)(x + 3)\).In the second fraction, the numerator \(3-x\) can be rewritten as \(-(x - 3)\), essentially factoring out \(-1\). The denominator \(x^3 + 3x^2\) can be factored by taking \(x^2\) common to get \(x^2(x + 3)\).

Step 2: Combine and Simplify

Now, multiply the two fractions together:\[\frac{(x + y)(x + 3)}{(x - 3)(x + 3)} imes \frac{-(x-3)}{x^2(x+3)} = \frac{-(x + y)(x + 3)(x-3)}{(x-3)(x + 3)x^2}\]Cancel out the common terms \((x - 3)\) and \((x + 3)\) from the numerator and denominator.

Step 3: Final Simplification

After cancelling, the expression simplifies to:\[\frac{-(x + y)}{x^2}\]

Key Concepts

Factoring Polynomials

The process of factoring polynomials involves breaking down a polynomial into simpler components, which are multiplied together to give the original expression. Imagine it as reverse distributing or finding common factors.
In the given exercise, the first step is to factor the numerator of the first fraction, \(x^{2} + 3x + xy + 3y\). This expression is factored by grouping, a technique used when an expression consists of four terms.

• First, identify pairs of terms that have common factors, \(x(x + 3)\) and \(y(x + 3)\).
• Next, factor by grouping each part to find a common factor: here it is \((x + 3)(x + y)\).

The denominator in the first fraction, \(x^{2} - 9\), is a classic example of a difference of squares, a type of factoring where \(a^{2} - b^{2}\) results in \((a - b)(a + b)\). Here, it simplifies to \((x - 3)(x + 3)\).
Factoring the second fraction requires understanding negative signs can be factored out: \(3-x\) becomes \(-(x-3)\), and the denominator \(x^3 + 3x^2\) simplifies to \(x^2(x + 3)\) by factoring out the greatest common factor \(x^2\). The key is recognizing patterns and simplifying each part to its basic components.

Simplifying Fractions

Once the numerator and denominator are factored, simplifying fractions simplifies them further by reducing them to their most basic form. To simplify a fraction, cancel out the common factors from the numerator and denominator.
Let's look at our product of fractions: \[\frac{(x + y)(x + 3)}{(x - 3)(x + 3)} \cdot \frac{-(x-3)}{x^2(x+3)}\]

• After factoring, the common terms in the fractions need to be cancelled. Here, \((x - 3)\) and \((x + 3)\) are found in both the numerator and the denominator, allowing us to cancel these terms out.
• The act of cancelling is essentially dividing both top and bottom by these common terms.

Doing this cancellation helps in reducing the expression to something simpler, which makes further operations easier. It's similar to reducing fractions to their lowest terms so they are more manageable to work with.

Multiplying Fractions

To multiply fractions, we need to multiply the numerators together and the denominators together then simplify the result if possible. It's a straightforward operation but benefits greatly from simplification before and after multiplying.
After cancelling common factors, the task was then to multiply the remaining terms:

• The numerator: \(-1 \times (x + y) = -(x + y)\)
• The denominator: \(x^2\)

Thus, the product simplifies to \(\frac{-(x + y)}{x^2}\).
One key point is ensuring you handle negative signs correctly, as they directly affect the final expression. It's also important to only cancel terms through multiplication rather than addition or subtraction. As you see here, the simplification is mainly achieved by identifying and cancelling the common factors before or during multiplication, which significantly simplifies the expression.